Solve x^2-1/1-x=5

$$\frac{ { x }^{ 2 } -1 }{ 1-x } = 5$$

Solve for x

$$x=-6$$

Steps Using Factoring

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$x^{2}-1=5\left(-x+1\right)$$

Use the distributive property to multiply 5 by -x+1.

$$x^{2}-1=-5x+5$$

Add 5x to both sides.

$$x^{2}-1+5x=5$$

Subtract 5 from both sides.

$$x^{2}-1+5x-5=0$$

Subtract 5 from -1 to get -6.

$$x^{2}-6+5x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+5x-6=0$$

To solve the equation, factor x^{2}+5x-6 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

Variable x cannot be equal to 1.

$$x=-6$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$x^{2}-1=5\left(-x+1\right)$$

Use the distributive property to multiply 5 by -x+1.

$$x^{2}-1=-5x+5$$

Add 5x to both sides.

$$x^{2}-1+5x=5$$

Subtract 5 from both sides.

$$x^{2}-1+5x-5=0$$

Subtract 5 from -1 to get -6.

$$x^{2}-6+5x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite x^{2}+5x-6 as \left(x^{2}-x\right)+\left(6x-6\right).

$$\left(x^{2}-x\right)+\left(6x-6\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-1\right)+6\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

Variable x cannot be equal to 1.

$$x=-6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$x^{2}-1=5\left(-x+1\right)$$

Use the distributive property to multiply 5 by -x+1.

$$x^{2}-1=-5x+5$$

Add 5x to both sides.

$$x^{2}-1+5x=5$$

Subtract 5 from both sides.

$$x^{2}-1+5x-5=0$$

Subtract 5 from -1 to get -6.

$$x^{2}-6+5x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-6\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-5±\sqrt{25+24}}{2}$$

Add 25 to 24.

$$x=\frac{-5±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-5±7}{2}$$

Now solve the equation x=\frac{-5±7}{2} when ± is plus. Add -5 to 7.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-5±7}{2} when ± is minus. Subtract 7 from -5.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=1 x=-6$$

Variable x cannot be equal to 1.

$$x=-6$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$x^{2}-1=5\left(-x+1\right)$$

Use the distributive property to multiply 5 by -x+1.

$$x^{2}-1=-5x+5$$

Add 5x to both sides.

$$x^{2}-1+5x=5$$

Add 1 to both sides.

$$x^{2}+5x=5+1$$

Add 5 and 1 to get 6.

$$x^{2}+5x=6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{7}{2} x+\frac{5}{2}=-\frac{7}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=1 x=-6$$

Variable x cannot be equal to 1.

$$x=-6$$